# Chapter_9_07 - Chapter 9 9.1 A 2.00 kg particle has the xy...

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Chapter 9 9.1 A 2.00 kg particle has the xy coordinates ( 1.20 m , 0.500 m ) , and a 4.00 kg particle has the xy coordinates (0.600 m , 0.750 m ). Both lie on a horizontal plane. At what (a) x and (b) y cooridatnes must you place a 3.00 kg particle such that the center of mass of the three particle system is ( 0.500 m , 0.700 m ) ? We begin by writing out the center of mass for x and y. We then solve for the unknown position of the 3kg mass. x cm = 2.00 kg ⋅ − 1.20 m + 4.00 kg 0.600 m + 3.00 kg x 2.00 kg + 4.00 kg + 3.00 kg = 2.00 kg ⋅ − 1.20 m + 4.00 kg 0.600 m + 3.00 kg x 9.00 kg = 3.00 kg x 9.00 kg x = 3 x cm = 1.5 m y cm = 2.00 kg 0.500 m + 4.00 kg ⋅ − 0.750 m + 3.00 kg y 2.00 kg + 4.00 kg + 3.00 kg = 2.00 kgm + 3.00 kg y 9.00 kg = 2.00 kgm 9.00 kg + 3.00 kg y 9.00 kg = 2.00 9.00 m + 1.00 3.00 y y = 3( y cm + 2.00 9.00 m ) = 3( 0.700 + 0.222) = 1.43 m 9.2 Figure 9-36 shows a three particle system. What are (a) the x coordinate and (b) the y coordinate of the center of mass of the three particle system . (c) What happens to the center of mass as the mass of the m 3 particle is increased. We can calculate the center of mass in the x and y directions.

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x cm = 1 ( m 1 + m 2 + m 3 ) ( m 1 x 1 + m 2 x 2 + m 3 x 3 ) = 1 (3.0 kg + 4.0 kg + 8.0 kg ) (3.0 kg 0 m + 4.0 kg 2.0 m + 8.0 kg 1.0 m ) = 1 (15 kg ) (16.0 kgm ) = 16 15 m y cm = 1 ( m 1 + m 2 + m 3 ) ( m 1 y 1 + m 2 y 2 + m 3 y 3 ) = 1 kg + 4.0 kg + 8.0 kg ) (3.0 kg 0 m + 4.0 kg 1.0 m + 8.0 kg 2.0 m ) = 1 kg ) (20.0 kgm ) = 20 15 m = 4 3 m As the topmost mass grows larger, the cm will move toward that mass. 9.3 What are (a) the x coordinate and (b) the y coordinate of the center of mass for the uniform plate shown in Fig 9-38 For continuous objects, we can often use symmetry to find the cm. In the drawing below, we have labeled the location of the center of each piece of the plate. We can now consider each piece of the plate as a point mass at the center. The mass of each plate is proportional to the area.
2L 4L 3L L (2L, 2.5L) (-L, -0.5L) (L, -3L) x y x cm = 1 Total Area Area i x i = (2 L 7 L ) ( L ) + (4 L L ) 2 L + (2 L 2 L ) L (2 L 7 L ) + (4 L L ) + (2 L 2 L ) = 2 L 3 22 L 2 = 1 11 L = 0.455 cm

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y cm = 1 Total Area Area i y i = (2 L 7 L ) ( 0.5 L ) + (4 L L ) 2.5 L + (2 L 2 L ) ( 3 L ) (2 L 7 L ) + (4 L L ) + (2 L 2 L ) = 9 L 3 22 L 2 = 9 22 L = 2.045 cm 9.7 The figure shows a slab with dimensions d 1 = 11.0 cm , d 2 = 2.80 cm , and d 3 = 11.0 cm . Half the slab consists of aluminum (density =2.70 g / cm 3 ) and half consists of iron density =7.85 g / cm 3 ) . What are the coordinates of the center of mass.
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## This note was uploaded on 02/10/2010 for the course BIOL 202 taught by Professor Bloom during the Spring '10 term at University of North Carolina Wilmington.

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Chapter_9_07 - Chapter 9 9.1 A 2.00 kg particle has the xy...

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